Rings of Polynomials

Donovan D Chang

24 Rings of Polynomials

We are used to thinking polynomials as an equation of a variable, say, $x$ . However, in this section, we will adopt a different viewpoint and an approach, and view polynomials in the context of a \textbf{ring}. To begin with, instead of referring $x$ as a variable, $x$ will be called an \textbf{indeterminate}, and our goal will be shifted to “finding a zero or zeroes of a polynomial” from “solving a polynomial equation”. Let us begin with a formal definition of a polynomial ring.

Definition. For a ring $R$ , a polynomial $f(x)$ with coefficients $a_i\in R$ is defined as an infinite formal sum as follows:

$\sum_{i=0}^{\infty}a_ix^i = a_0+a_1x+\cdots +a_nx^n+\cdots$

where $a_i=0$ except for a finite number of $i$ ‘s.

As usual, $a_i$ ‘s are the coefficients of $f(x)$ , and the largest value of $i$ such that $a_i\neq 0$ is the degree of $f(x)$ .

We briefly introduce the underlying theory behind “solving a polynomial equation”, the evaluation homomorphisms for field theory [9]. Though seemingly trivial, in fact this is of crucial importance for the rest of our work.

Theorem. (Evaluation Homomorphisms for Field Theory)
Suppose $F\leq E$ , where $F$ is a field, and let $\alpha\in E$ .
Then, the map $\phi_\alpha:F[x]\rightarrow E$ defined by
$\phi_\alpha \left( a_0+a_1x+\cdots+a_nx^n\right)</em> <em>= a_0+a_1\alpha+\cdots+a_n\alpha^n$
for $\phi_\alpha \left( a_0+a_1x+\cdots+a_nx^n\right)\in F[x]$ is a homomorphism.
Also, $\phi_\alpha(x)=\alpha$ , and $\phi_\alpha(a)=a$ for all $a\in F$ by $I:\phi_\alpha(a)\rightarrow\phi_\alpha(a)$ .

Now, we can simply input, or substitute, our indeterminate $x$ with numbers of our interest. Owing to the evaluation homomorphisms for field theory, our journey of finding a zero of a polynomial has become finding

$f(\alpha)=0$

such that

$\phi_\alpha \left( f(x)\right) = a_0+a_1\alpha+\cdots+a_n\alpha^n$

If we let $F=\mathbb{Q}$ and $E=\mathbb{R}$ , the evaluation homomorphism for field theory turns into a common example found in elementary algebra.

Factoring Over a Field

We begin with the division algorithm for $F[x]$ [9] as this is the very basic tool of factorizing polynomials over a field.

Theorem. (Division Algorithm for $F[x]$ )
Let $f(x),g(x)\in F[x]$ with degrees $n$ and $m$ , respectively, where $n\geq m>0$ . Then, there exist unique polynomials $q(x)$ and $r(x)$ such that

$f(x) = g(x)q(x)+r(x)$

where $r(x)=0$ or the degree of $r(x)$ is smaller than $m$ .

Let us illustrate with an example [9].

Example. Find the factorization of a polynomial $x^2+4x+3$ in $\mathbb{Z}_7[x]$ .

Let $f(x) = x^2+4x+3$ .
Then, $f(4) = 16+16+3\equiv 0$ (mod 7).
Therefore, $x-4=x+3$ is a factor of $f$ .

Also, $f(6) = 36+24+3\equiv 0$ (mod 7).

Therefore, $x-6=x+1$ is a factor of $f$ .

Therefore, $x^2+4x+3=(x+1)(x+3)$ .

Irreducible Polynomials

Not all polynomials can be factorized. Such polynomials are referred to as irreducible polynomials}, and formally defined [9] as:

Definition. (Irreducible Polynomials)
Let $f(x),g(x),h(x)\in F[x]$ .
We say $f(x)$ , a polynomial with the first or greater degree, is irreducible over $F$ or is an irreducible polynomial in $F[x]$ , if there does not exist $g(x)$ and $h(x)$ such that

$f(x)=g(x)h(x)$

where the degrees of $g(x)$ and $h(x)$ are both lower than the degree of $f$ .

Note that a polynomial can be either reducible or irreducible depending on the context. For example, $x^2-2$ is reducible in $\mathbb{R}[x]$ as

$\left(x-\sqrt{2}\right) \left(x+\sqrt{2}\right)$

yet is irreducible in $\mathbb{Q}[x]$ , since $\sqrt{2}\notin \mathbb{Q}$ .

Let us illustrate with an example.

Example. Demonstrate the irreducibility of $x^4-22x+1$ over $\mathbb{Q}$ .

We shall prove by contradiction, and as such, let us suppose the opposite, that $f(x)=x^4-22x+1$ is reducible over $\mathbb{Q}$ , where $f(x)=x^4-22x+1$ .\\
Then, $f$ has either a zero, or is reduced to two quadratic polynomials.

$f(x)$ has a zero in $\mathbb{Q}$ .
Then, by corollary 23.12[9], the zero $m\in \mathbb{Z}$ , and divides 1.
Then, $m\in \lbrace -1,1\rbrace$ .
However, $f(1)=f(-1)=1-22+1=-20\neq 0$ .
Therefore, $f(x)$ does not have a zero in $\mathbb{Q}$ .
is reduced to two quadratic polynomials.
Then, by theorem 23.11 [9], there exists such that , where the order of and is 2.By letting and , and matching the coefficients of with , there are two cases as follows:
1. $b=d=1$
  Then, $a+c=0$ and $ac=-24$ .
  However, there does not exist $a,c\in \mathbb{Z}$ satisfying the equations above.
  Therefore, there does not exist $g(x),h(x)\in \mathbb{Z}[x]$ such that $g(x)h(x)=f(x)$ , where the order of $g$ and $h$ is 2.
2. $b=d=-1$
  Then, $a+c=0$ and $ac=-20$ .
  Again, there does not exist $a,c\in \mathbb{Z}$ satisfying the equations above.
  Therefore, there does not exist $g(x),h(x)\in \mathbb{Z}[x]$ such that $g(x)h(x)=f(x)$ , where the order of $g$ and $h$ is 2.

Therefore, $f$ neither has a zero nor is reducible to 2 quadratic polynomials.
This is a contradiction.

Therefore, $f$ is not irreducible over $\mathbb{Q}$ .

We conclude this section introducing the theorem about the uniqueness of factorization in $F[x]$ [9]. While a full proof is not provided, the uniqueness should come intuitive to readers.

Theorem. (Uniqueness of Factorization in $F[x]$ )
When polynomials in $F[x]$ are factorized until no more factorization is possible, the irreducible polynomials are unique except for order and non-zero constant factors.

24 Rings of Polynomials

Factoring Over a Field

Irreducible Polynomials

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